CURVES WHOSE SECANT DEGREE IS ONE IN POSITIVE CHARACTERISTIC. 1. Introduction


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1 Acta Math. Univ. Comenianae Vol. LXXXI, 1 (2012), pp CURVES WHOSE SECANT DEGREE IS ONE IN POSITIVE CHARACTERISTIC E. BALLICO Abstract. Here we study (in positive characteristic) integral curves X P r with secant degree one, i.e., for which a general P Sec k 1 (X) is in a unique ksecant (k 1)dimensional linear subspace. 1. Introduction Let K be an algebraically closed base field. Let X P r be an integral and nondegenerate closed subvariety. For each x {0,..., r}, let G(x, r) denote the Grassmannian of all xdimensional linear subspaces of P r. For each integer k 1 let σ k (X) denote the closure in P r of the union of all A G(k 1, r) spanned by k points of X (the variety σ k (X) is sometimes called the (k 1)secant variety of X and written Sec k 1 (X), but we prefer to call it the ksecant variety of X). The integral variety σ k (X) may be obtained in the following way. Assume that X is nondegenerate. For any closed subscheme E P r let E denote its linear span. Let V (X, k) G(k 1, r) denote the closure in G(k 1, r) of the set of all A G(k 1, r) spanned by kpoints of X. Set S[X, k] := {(P, A) P r G(k 1, r): P A, A V (X, k)}. Let p 1 : P r G(k 1, r) P r denote the projection onto the first factor. We have σ k (X) = p 1 (S[X, k]). Set m X,k := p 1 S[X,k]. If σ k (X) has the expected dimension k (dim(x) + 1) 1 (i.e., if m X,k is generically finite), then we write i k (X) for the inseparable degree of m X,k and s k (X) for its separable degree. For any P X reg, let T P X P r denote the tangent space to X at P. If k 2, we say that X is kunconstrained if dim( T P1 X T Pk X ) = dim(σ k (X)) for a general (P 1,..., P k ) X k. Terracini s lemma says that dim( T P1 X T Pk X ) dim(σ k (X))) Received May 22, 2011; revised November 3, Mathematics Subject Classification. Primary 14N05; 14H50. Key words and phrases. Join; secant variety; rational normal curve. The author was partially supported by MIUR and GNSAGA of INdAM (Italy).
2 72 E. BALLICO and that in characteristic zero equality always holds ([1, 1] or [3, 2]). The case k = 2 of this notion was introduced in [3]. A nondegenerate curve Y P r is 2unconstrained if and only if either r = 2 or Y is not strange [3, Example (e1) at page 333]. From now on we assume dim(x) = 1. We first prove the following result. Theorem 1. Fix integers r 2k 4. Let X P r be an integral, nondegenerate and kunconstrained curve. Then s k (X) = 1. For each integer i such that 2 2i r we define the integer e i (X) in the following way. Fix a general (P 1,..., P i ) X i. Thus P j X reg for all j. Set V := T P1 X T Pi X. Notice that (V X) red {P 1,..., P i } and the scheme V X is zerodimensional. Varying (P 1,..., P i ) in X i we see that each P j appears with the same multiplicity in the zerodimensional scheme V X. We call e i (X) this multiplicity. In characteristic zero we always have e i (X) = 2. The integer e 1 (X) is the intersection multiplicity of X with its general tangent line at its contact point. Hence if char(k) is odd the curve X is reflexive if and only if e 1 (X) = 2 ([4, 3.5]). In the general case we have e 1 (X) 2 and e i (X) e i+1 (X). For any P X reg and any integer t {1,..., r}, let O(X, P, t) G(t, r) denote the tdimensional osculating plane of X at P. Thus O(X, P, 1) = T P X. Fix integers i 1, and j h 0, 1 h i. We only need the case 2i+ i h=1 j h r. Fix a general (P 1,..., P i ) X i and set V := i h=1 O(X, P h, 1 + j h ). For any h {1,..., i}, let E(X; i; j 1,... j i ; h) be the multiplicity of P h in the scheme V X. We will only use the case j 1 = 1 and j h = 0 for all h 1. If either char(k) = 0 or char(k) > deg(x), then E(X; i; j 1,... j i ; h) = 2 + j h (Lemma 9). Here we prove the following result. Theorem 2. Let X P 2k 1, k 2, be an integral, nondegenerate and k unconstrained curve. Set j 1 := 1 and j h := 0 for all h {2,..., k 1}. (a) If s k (X) = 1 and E(X; k 1; j 1,..., j k 1 ; 1) = e k 1 (X) + 1, then X is smooth and rational and deg(x) = (k 1)e k 1 (X) + 1. (b) X is a rational normal curve if and only if s k (X) = 1, e k 1 (X) = 2 and E(X; k 1; j 1,..., j k 1 ; 1) = 3. We do not know if in the statement of Theorem 2 we may drop the conditions e k 1 (X) = 2 and E(X; k 1; j 1,..., j k 1 ; 1) = 3. We are able to prove that we may drop the first one in the case k = 2, i.e., we prove the following result. Proposition 1. Let X P 3 be an integral and nondegenerate curve. following conditions are equivalent: (a) X is not strange, s 2 (X) = 1 and E(X; 1; 1; 1) = e 1 (X) + 1; (b) i 2 (X) = s 2 (X) = 1 and E(X; 1; 1; 1) = e 1 (X) + 1; (c) X is a rational normal curve. The picture is very easy if char(k) > deg(x). As a byproduct of Theorem 2 we give the following result. Theorem 3. Let X P 2k 1 be an integral and nondegenerate curve. Assume char(k) > deg(x). X is a rational normal curve if and only if s k (X) = 1. The
3 SECANT VARIETIES The proofs Remark 1. Assume X of arbitrary dimension and that dim(σ k (X)) = k(dim(x) + 1) 1. As in [3] (the case k = 2) X is kunconstrained if and only if i k (X) = 1. Lemma 1. Fix integers c > 0, s > y 2 and r s(c + 1) 1. Let X P r be an integral and nondegenerate cdimensional subvariety such that dim(σ s (X)) = s(c + 1) 1. If X is sunconstrained, then X is yunconstrained. Proof. Since dim(σ s (X)) = s(c + 1) 1 and X is sunconstrained, we have dim( T P1 X T Ps X = s(c + 1) 1 for a general (P 1,..., P s ) X s. Hence dim( T P1 X T Py (X) = y(c + 1) 1. Hence X is yunconstrained. We recall the following very useful result ([1, 1]). Lemma 2. Let X P r be an integral and nondegenerate curve. Then X is nondefective, i.e., dim(σ a (X)) = min{r, 2a 1} for all integers a 2. From Lemmas 1 and 2 we get the following result. Lemma 3. Fix integers s > y 2 and r 2s 1. Let X P r be an integral and nondegenerate curve. If X is sunconstrained, then X is yunconstrained and not strange. We recall that a finite set S P x is said to be in linearly general position if dim( S ) = min{x, (S ) 1} for every S S. The general hyperplane section of a nondegenerate curve X P r is in linearly general position if X is not strange ([6, Lemma 1.1]). Hence Lemma 3 implies the following result. Lemma 4. Fix integers r, s such that r 2s 1 3. Let X P r be an integral and nondegenerate curve. Assume that X is sunconstrained. Then X is not strange and a general hyperplane section of X is in linearly general position. Proof of Theorem 1. We extend the proof of the case k = 2 given in [3, 4]. By Lemma 4 a general (k 1)dimensional ksecant plane of X meets X at exactly k points. Fix a general (P 1,..., P k ) X k and set V := T P1 X T Pk. Since X is kunconstrained, we have dim(v ) = 2k 1. Since 2k 1 < r and X is nondegenerate, the set S := (V X) red is finite. Fix a general P {P 1,..., P k }. Assume s k (X) 2. Since a general hyperplane section of X is in linearly general position (Lemma 4), the integer s k (X) is the number of different kples of points of X such that a general point of σ k (X) is in their linear span. Since P may be considered as a general point of σ k (X) and s k (X) 2, there is (Q 1,..., Q k ) X k such that P {Q 1,..., Q k } and {P 1,..., P k } {Q 1,..., Q k }. For general P we may also assume that (Q 1,..., Q k ) is general in X k. Hence each P i and each Q j is a smooth point of X. Terracini s lemma gives T P1 X T Pk X T P σ k (X) and T Q1 X T Qk X T P σ k (X). Since X is kunconstrained and both (P 1,..., P k )
4 74 E. BALLICO and (Q 1,..., Q k ) are general in X k, we have T P1 X T Pk = T P σ k (X) and T Q1 X T Qk X = T P σ k (X). Hence {Q 1,..., Q k } S. Since S is finite, the union of the linear spans of all S S with (S ) = k is a finite number of linear subspaces of dimension at most k 1 and S = {P 1,..., P k } if and only if S = {P 1,..., P k }, because {P 1,..., P k } X = {P 1,..., P k }. Hence dim( S {P 1,..., P k } ) k 2 for all S {P 1,..., P k }. Varying P {P 1,..., P k } = P k 1, we get a contradiction. Lemma 5. Let X P r, r 2k 1 5, be an integral, nondegenerate and kunconstrained curve. Fix an integer s such that 1 s k 2. Fix a general (A 1,..., A s ) X s and set W := T A1 X T As X. Then dim(w ) = 2s 1. Let l W : P r \ W P r 2s denote the linear projection from W. Let Y P r 2s denote the closure of l W (Y \ Y W ). Then Y is (k s)unconstrained and it is not strage. Proof. Fix a general A s+1,..., A k X k s. Notice that (l W (A s+1 ),..., l W (A k )) is general in Y k s and l W ( W T As+1 X T Ak X \ W ) = T lw (A s+1)y T lw (A k ))Y. Hence the latter space has dimension 2k 2s 1. Hence Y is (k s)unconstrained. Since k s 2, Y is not strange. Lemma 6. Fix integers c > 0, k 2 and r (c + 1)k 1. Let X P r be a kunconstrained cdimensional variety such that dim(σ k (X)) = (c + 1)k 1. Fix an integer s {1,..., k 1} and a general (P 1,..., P s ) X s. Set V := T P1 X T Ps X. Then dim(v ) = (c + 1)s 1 and the restriction to X of the linear projection l V : P r \V P r (c+1)s is a generically finite separable morphism. Proof. Since s + 1 k and dim(σ k (X)) = (c + 1)k 1, we have dim(σ s (X)) = (c + 1)s 1. Lemma 1 gives that X is sunconstrained. Since X is (s + 1)unconstrained and dim(σ s+1 (X)) = (c + 1)(s + 1) 1, we have dim( V T P X ) = dim(v ) + dim(t P X) + 1 for a general P X, i.e., V T P X = for a general P X. Thus l V (X \ V ) has differential with rank c, i.e., it is separable and generically finite. Proof of Theorem 2. If X is a rational normal curve, then it is kunconstrained, s k (X) = 1 ([2, First 4 lines of page 128]) and i k (X) = 1 (Remark 1). Now assume s k (X) = 1. In step (c) we will use the assumption E(X; k 1; 1, 0,..., 0; 1) = e k 1 (X) + 1. We need to adapt a part of the characteristic zero proof given in [2] to the positive characteristic case. We will follow [2] as much as possible. Fix a general (P 1,..., P k 1 ) X k 1 and set V := T P1 X T Pk 1 X. Since X is kunconstrained, we have dim(v ) = 2k 3. Since X is nondegenerate, the set S := (V X) red is finite. (a) Here we check that S X reg. If k = 2, then for a general P 1 we have T P1 X Sing(X) =, because X is not strange by [3, Example (e1) at page 333]. Now assume k 3. Since X is not strange (use Lemma 1), for general P 1 X, we have T P1 X Sing(X) =. Then by induction on i we check using a linear
5 SECANT VARIETIES 75 projection from T Pi X as in Lemma 5 that T P1 X T Pi X Sing(X) = (more precisely, for any finite set Σ X we check by induction on i that T P1 X T Pi X Σ = for a general (P 1,..., P i ) X i ). For i = k 1 we get S X reg. (b) Obviously {P 1,..., P k 1 } S. Here we check that S = {P 1,..., P k 1 }. Assume for the moment the existence of Q S \ {P 1,..., P k 1 }. Since X is not strange, it is not very strange, i.e., a general hyperplane section of X is in linearly general position ([6, Lemma 1.1]). Since (P 1,..., P k 1 ) is general in X k 1, we get {P 1,..., P k 1 } X = {P 1,..., P k 1 }. Thus dim( {P 1,..., P k 1, Q} ) = k 1. Fix a general z {P 1,..., P k 1, Q}. We have P 2k 1 = T z σ k (X) T P1 X T Pk 1 X T Q X (Terracini s lemma ([3, 2] or [1, Proposition 1.9]). The additive map giving Terracini s lemma for joins in the proof of [1, Proposition 1.9], shows that the map m X,k has noninvertible differential over the point z. Since P 2k 1 is smooth and m X,k is separable, we get that m X,k is not finite of degree 1 near z. Since s k (X) = 1, m X,k contracts a curve over z. Since z lies in infinitely many (k 1)dimensional ksecant subspaces, we get that dim(σ k (X)) 2k 2, contradicting Lemma 2. The contradiction proves S = {P 1,..., P k 1 }. (c) Step (b) means that {P 1,..., P k 1 } is the reduction of the scheme theoretically intersection X V. Let Z i denote the connected component of the scheme X V supported by P i. Set e := deg(z 1 ). Since T P1 X V, we have e 2. Varying (P 1,..., P k 1 ) in X k 1 we get deg(z i ) = e for all i. The definition of the integer e k 1 (X) gives e = e k 1 (X). Set φ := l V (X \ V X). Since X V X reg, φ is dominant and X reg is a smooth curve, φ induces a finite morphism ψ : X P 1. Bezout s theorem gives deg(x) = (k 1)e + deg(ψ). Lemma 6 gives that ψ is separable. Hence deg(ψ) is the separable degree of ψ. Assume deg(ψ) 2. Since P 1 is algebraically simply connected, there is Q X at which ψ ramifies. First assume Q X reg. Since E(X; k 1; 1, 0,..., 0; 1) = e k 1 (X) + 1, ψ is not ramified at P 1. Moving P 1,..., P k 1 we get Q / {P 1,..., P k 1 }. The definition of φ gives dim(v T Q X) dim(v ) + 1. Hence the additive map giving Terracini s lemma for joins in the proof of [1, Proposition 1.9], shows that the map m X,k has noninvertible differential over the general point z {P 1,..., P k 1, Q}. As in step (b) we get a contradiction. Now assume Q Sing(X). Let u: C X denote the normalization map. Since we assumed deg(ψ) 2, we have deg(ψ u) 2. Since P 1 is algebraically simply connected, there is Q C such that ψ u is ramified at Q. We repeat the construction of joins and secant variety starting from the nonembedded curve C and get a contradiction using Q instead of Q. Thus deg(ψ) = 1, i.e. deg(x) = (k 1)e k 1 (X) + 1, and X is rational. X is a rational normal curve if and only if deg(x) = 2k 1, i.e., if and only if e = 2. Take any P Sing(X) (if any). Set H := {P } V. Since X is singular
6 76 E. BALLICO at P, we have deg(h X) 2 + (k 1)e > deg(x), that is contradiction. Thus X is smooth. Proof of Proposition 1. We have i 2 (X) = 1 if and only if X is 2unconstrained ([3] or Remark 1). Obviously X is 2unconstrained. Hence it is sufficient to prove that if X is 2unconstrained, s 2 (X) = 1, and E(X; 1; 1; 1) = e 1 (X) + 1, then X is a rational normal curve. Theorem 2 says that X is smooth and rational and deg(x) = e 1 (X) + 1. Thus it is sufficient to prove e 1 (X) = 2. Assume e 1 (X) 3. Since deg(x) = e 1 (X) + 1, Bezout s theorem says that any two different tangent lines are disjoint. Let T X P 3 denote the tangent developable of X. Fix a general P P 3 and let l P : P 3 \ {P } P 2 be the linear projection from P. Set l := l P X. Since P / T X, l is unramified. Since X is smooth, s 2 (X) = 1 and P is general, the map l is birational onto its image and the curve l(x) has a unique singular point (the point l(p 1 ) = l(p 2 ) with P {P 1, P 2 } and (P 1, P 2 ) X 2 ). We have p a (l(x)) = e 1 (X)(e 1 (X) 1)/2 2. Since P / T X, we have P / T Pi X, i = 1, 2. Since T P1 X T P2 (X) =, the line T P2 X is not contained in the plane {P } T P1 X. Thus l P (T P1 X) l P (T P2 X). Thus l(p 1 ) is an ordinary double point of l(x). Hence l(x) has geometric genus p a (X) 1 > 0, thath is contradiction. Lemma 7. Let X P r be an integral and nondegenerate curve. Assume char(k) > deg(x). Then e i (X) = 2 for all positive integers i such that 2i r. Proof. We have e 1 (X) = 2, because in large characteristic the Hermite sequence of X at its general point is the classical one ([5, Theorem 15]). The case i 2 is obtained by induction on i taking instead of X its image by the linear projection from T Pi X, P i general in X. Lemma 8. Let X P r be an integral and nondegenerate curve. Assume char(k) > deg(x). Then X is iunconstrained for all integers i 2. Proof. Fix a linear subspace V P r such that v := dim(v ) r 2. Let l V : P r \ V P r v 1 denote the linear projection from V. Since char(k) > deg(x), the restriction of l V to X is separable. Hence T Pi X V = for a general P i X. Take V = T P1 X T Pi 1 X with (P 1,..., P i 1 ) general in X i 1 and use induction on i. Lemma 9. Let X P r be an integral and nondegenerate curve. Assume char(k) > deg(x). Then E(X; i; j 1,..., j i ; h) = 2 + j h for all i, j 1,..., j i such that i 2i + j x r and for a general (P 1,..., P i ) X i, the linear span of the osculating spaces has dimension 2i 1 + i x=1 j x. x=1 O(X, P x, 1 + j x ), 1 x i,
7 SECANT VARIETIES 77 Proof. The case i = 1 is true by [5, Theorem 15]. Hence we may assume i 2. Fix an index c {1,..., i} \ {h}. For a general P c X, the point P c appears with multiplicity exactly j c + 2 in the scheme O(X, P c, j c + 1) ([5, Theorem 15]). Since char(k) > deg(x), the rational map l obtained restricting to X the linear projection from O(X, P c, 1 + j c ) is separable. Call Y the closure in P r jc 2 of l(x \ O(X, P, 1 + j c ) X). Take P x, x c, such that (P 1,..., P i ) is general in X c and write Q x := l(p x ) for all x c. Let V be the linear span of the osculating spaces O(X, P x, 1 + j x ), 1 x i, U the linear span of the osculating spaces O(X, P x, 1 + j x ), x c, and W the linear span of the osculating spaces O(Y, Q x, 1 + j x ), x c. By the inductive assumption U and W have dimension 2i 3 + x c j x. Hence l(u) = W and dim(v ) = 2i 1 + i x=1 j x. Since the points Q i are general and l is separable, the scheme l 1 ((2 + j x )Q x )), x c, is a divisor of X whose connected component supported by P x has degree 2 + j x. Use the inductive assumption on Y to get E(X; i; j 1,..., j i ; h) = 2 + j h. Proof of Theorem 3. Apply Theorem 2 and Lemmas 7, 8 and 9. We want to thank the referee for her/his useful observa Acknowledgment. tions. References 1. Ådlandsvik B., Joins and higher secant varieties, Math. Scand. 61 (1987), CatalanoJohnson M., The homogeneous ideal of higher secant varieties, J. Pure Appl. Algebra 158 (2001), Dale M., Terracini s lemma and the secant variety of a curve, Proc. London Math. Soc. 49(3) (1984), Hefez A. and Kleiman S. L., Notes on the duality of projective varieties. Geometry today (Rome, 1984), , Progr. Math., 60, Birkhäuser Boston, Boston, MA, Laksov D., Wronskians and Plücker formulas for linear systems on curves, Ann. Scient. École Norm. Sup. 17 (1984), Rathmann J., The uniform position principle for curves in characteristic p, Math. Ann. 276(4) (1987), E. Ballico, Department of Mathematics, University of Trento, Povo (TN), Italy,
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